A chemist prepares 1.0 L of a buffer solution by preparing an aqueous solution containing 1.00 mol of H2PO4−, then adding 0.27 moles of KOH to convert the acid into its conjugate base. What is the final pH of the solution? The Ka for H2PO4− is 6.2 x 10−8; the pKa is 7.20.

pH of solution is 6.77.

Explanation:

H₂PO₄⁻ reacts with KOH thus:

H₂PO₄⁻ + KOH → HPO₄²⁻ + H₂O + K⁺

If 1.00 mol of H₂PO₄⁻ reacts with 0.27mol of KOH, moles of HPO₄²⁻ produced are 0.27mol and moles of H₂PO₄⁻ that remains are 1.00mol - 0.27mol = 0.73mol.

It is possible to find pH of this buffer using H-H equation:

pH = pka + log₁₀ [HPO₄²⁻] / [H₂PO₄⁻]

Where pKa is 7.20

[HPO₄²⁻] = 0.27mol / 1.0L = 0.27M

[H₂PO₄⁻] = 0.73mol / 1.0L = 0.73M

Replacing:

pH = 7.20 + log₁₀ [0.27M] / [0.73M]

pH = 6.77

pH of solution is 6.77

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.